[PHP] Validating form field text input to be a specific variable type
All,
I have a form on which the user is supposed to select a variable type
(boolean, integer, real, date/time, text) from a select box and enter the
default value for this selected variable type in a text box. I'm trying to
validate that the default value entered matches the variable type selected
i.e. user selects boolean so valid defaults could only be 0, 1, true, false
and anything else would generate an error, or the user selects integer and
enters 1.7 for the default would also throw a flag. I know that all of the
form values submitted from the web page are strings but is there a way to
test/convert the strings against the other variable types.
I'm sure that I'm not explaining this very well, but for example, if I use
the following code I will always get an error displayed even if the user
enters a valid value such as 3 in the default field because the form values
are always submitted as strings.
if ( ($GLOBALS['PageOptions']['Type'] == 'Integer') and (!
is_int($GLOBALS['PageOptions']['Default']) ) )
{
// display an error
}
If I use (int) on the form default value then that won't work either because
if the default field value entered was not an integer but text such as
'snafu' then the value is always converted to an integer regardless.
if ( ($GLOBALS['PageOptions']['Type'] == 'Integer') and (! is_int((int)
$GLOBALS['PageOptions']['Default']) ) )
{
// display an error
}
Thanks in advance,
Dave Merritt
[EMAIL PROTECTED]
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RE: [PHP] Validating form field text input to be a specific variable type
Okay seems to makes sense, but when I do the following it doesn't appear to
be working correctly, or I'm viewing my logic incorrectly one:
if ( (int)$PageOptions['Default'] == $PageOptions['Default'] )
{
// display a message if the integer of the default value matches the
default value
}
Enter 5, message displayed, so correct
Enter 5.0, messaged displayed, incorrect
Enter 5.5, no message, so correct
Enter a, message displayed, incorrect
Enter 5.0a, message displayed, incorrect
Enter 5.5a, no message, so correct
Enter a5.5, message displayed, incorrect
What am I missing here?
Thanks,
Dave Merritt
[EMAIL PROTECTED]
-Original Message-
From: John W. Holmes [mailto:[EMAIL PROTECTED]
Sent: Tuesday, April 06, 2004 3:19 PM
To: Merritt, Dave; [EMAIL PROTECTED]
Subject: Re: [PHP] Validating form field text input to be a specific
variable type
From: Merritt, Dave [EMAIL PROTECTED]
I have a form on which the user is supposed to select a variable type
(boolean, integer, real, date/time, text) from a select box and enter
the default value for this selected variable type in a text box. I'm
trying
to
validate that the default value entered matches the variable type
selected i.e. user selects boolean so valid defaults could only be 0,
1, true,
false
and anything else would generate an error, or the user selects integer
and enters 1.7 for the default would also throw a flag. I know that
all of
the
form values submitted from the web page are strings but is there a way
to test/convert the strings against the other variable types.
I'm sure that I'm not explaining this very well, but for example, if I
use the following code I will always get an error displayed even if
the user enters a valid value such as 3 in the default field because
the form
values
are always submitted as strings.
Well, if (int)$string == $string, then the value is an integer. Same for
(float)$string == $string for a real number. Boolean would be easy, just
strtolower($string) as compare to 1, 0, 'true', or 'false'. Date/time
validation will probaby require a regular expression or breaking it up to
validate days/month, etc. That can get a little hairy. If they say text,
well, anything goes, right? Maybe just make sure it's not empty()?
Let me know if you need more details. There are probably a ton of different
ways to do this.
---John Holmes...
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RE: [PHP] Validating form field text input to be a specific variable type
Okay seems to makes sense, but when I do the following it doesn't appear to
be working correctly, or I'm viewing my logic incorrectly one:
if ( (int)$PageOptions['Default'] == $PageOptions['Default'] ) {
// display a message if the integer of the default value matches the
default value }
Enter 5, message displayed, so correct
Enter 5.0, messaged displayed, incorrect
Enter 5.5, no message, so correct
Enter a, message displayed, incorrect
Enter 5.0a, message displayed, incorrect
Enter 5.5a, no message, so correct
Enter a5.5, message displayed, incorrect
What am I missing here?
Thanks,
Dave Merritt
[EMAIL PROTECTED]
-Original Message-
From: John W. Holmes [mailto:[EMAIL PROTECTED]
Sent: Tuesday, April 06, 2004 3:19 PM
To: Merritt, Dave; [EMAIL PROTECTED]
Subject: Re: [PHP] Validating form field text input to be a specific
variable type
From: Merritt, Dave [EMAIL PROTECTED]
I have a form on which the user is supposed to select a variable type
(boolean, integer, real, date/time, text) from a select box and enter
the default value for this selected variable type in a text box. I'm
trying
to
validate that the default value entered matches the variable type
selected i.e. user selects boolean so valid defaults could only be 0,
1, true,
false
and anything else would generate an error, or the user selects integer
and enters 1.7 for the default would also throw a flag. I know that
all of
the
form values submitted from the web page are strings but is there a way
to test/convert the strings against the other variable types.
I'm sure that I'm not explaining this very well, but for example, if I
use the following code I will always get an error displayed even if
the user enters a valid value such as 3 in the default field because
the form
values
are always submitted as strings.
Well, if (int)$string == $string, then the value is an integer. Same for
(float)$string == $string for a real number. Boolean would be easy, just
strtolower($string) as compare to 1, 0, 'true', or 'false'. Date/time
validation will probaby require a regular expression or breaking it up to
validate days/month, etc. That can get a little hairy. If they say text,
well, anything goes, right? Maybe just make sure it's not empty()?
Let me know if you need more details. There are probably a ton of different
ways to do this.
---John Holmes...
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[PHP] combining multi-dimensional arrays from multiple parse_ini_file results
All,
How do I combine the multi-dimensional arrays returned from the
parse_ini_file function on two different ini files? I tried:
$my_array = parse_ini_file('some_file', true);
$my_array[] = parse_ini_file['another_file', true);
Thanks
Dave
[EMAIL PROTECTED]
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[PHP] To use PEAR::DB or the PHP database functions
All, I've always used MySQL databases and the MySQL functions in PHP for my code in the past. However, I'm now working on a project that I want the project to be able to be database independent so that the user of the project can use whatever database he/she wishes. I'm looking primarily at providing support for MySQL, PostgreSQL, Oracle, SQL Server databases. What's the general consensus on how to handle this? Do I need to look at using PEAR::DB so that the type of database is hidden from my code or would I look at writing different include files for each database type and each of the include files use the relevant PHP functions? Or some other totally different way? Thanks Dave Merritt [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Recommendations for PHP/MySQl calendar on-line submission/voting
All, I working with helping our local high school technical center set up a web server/site. Went with the typical AMP solution (unfortunately on MSWin though). The center is wanting to add an on-line calendar system. This doesn't need to be anything fancy, just something that will display a monthly view with event titles, a way to provide detail content for the event, and a way for multiple users to add/edit/remove entries. I'm doing some searching for solutions but there's definitely plenty to choose from., so if anyone has any recommendations I'd appreciate it. Also, the school tries to get a lot of community involvement by doing a lot of Name the ... whatever the activity is at time competitions and they are wanting to come up with a way to do online entry submission and online voting/polling of the entries. Obviously I could write this, but was wondering if anybody had any suggestions of existing code to do this. TIA Dave Merritt [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] How to show autocad (dwf, dwg) files in browser?
Lars, You say you downloaded the Volo View Express, but your plug-in page points to the WHIP viewer plug. Regardless, there are two versions of Volo View. Did you download the DWF version only of Volo View? Dave -Original Message- From: Lars Espelid [mailto:lars_espelid;hotmail.com] Sent: Friday, November 15, 2002 4:09 PM To: [EMAIL PROTECTED] Subject: [PHP] How to show autocad (dwf, dwg) files in browser? HI, I have downloaded Volo View Express from www.autodesk.com (plugin to web-browser) and can see the sample autocad-file that is presented on their site. Code ment to display the file drawing1.dwg on my page (none working): 1) img src=drawing1.dwg Displays no image. 2) OBJECT data=drawing1.dwg type=image/vnd.dwg A nice drawing. /OBJECT The only thing I can see is a blue rectangle whith the following error message inside it: Drawing File Format Unrecogr 3) object param name=Filename value=drawing1.dwg embed name=thanks src=drawing1.dwg pluginspage=http://www.autodesk.com/whip; /object The only thing I can see is a blue rectangle whith the following error message inside it: Drawing File Format Unrecogr I'm running Apache 1.3.26. This is what I have written into httpd.conf: AddType model/vnd.dwf .dwf #drawing/x-dwf .dwf AddType image/vnd.dxf .dxf AddType image/vnd.dwg .dwg This is what I have written into mime.types and mime.types.default: image/vnd.dwg image/vnd.dxf Appreciate any help. Thanks Lars -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] RE: -- OT -- [PHP] How to show autocad (dwf, dwg) files in browser?
Okay, the WHIP viewer is only for DWF files from ACAD 13, 14, 2000 only. It does NOT handle DWG files or DWF from ACAD 2002 based products (Inventor), Volo View Express 2.01 handles DWF, DWG, DXF only if you download the english language version. Non-english versions of VVE will NOT view DWG DXF files only DWF files from ACAD 2002. What version fo ACAD are you dealing with? If it' 13, 14, or 2000 or going to have to use WHIP DWF files. If it's ACAD 2002 and you can use an english language viewer then you can use DWF, DWG, or DXF. If it's ACAD 2002 and your using a non-english viewer then it's DWF only. If you have a site that has a mixture of ACAD versions from 13 thru 2002, or going to have to have to either distinguish between the versions somehow and use the appropriate viewer or you will have to open each file into ACAD 2002 and save each one back into 2002 format. The file format for DWF changed between ACAD 2000 2002. Sucks I know, esp. the english/non-english thing. Dave -Original Message- From: Lars Espelid [mailto:lars_espelid;hotmail.com] Sent: Friday, November 15, 2002 6:09 PM To: [EMAIL PROTECTED] Subject: Re: [PHP] How to show autocad (dwf, dwg) files in browser? Only my third code-example points to the whip-viewer plug, and that's most likely wrong. My first and second example may also be wrong, someone know how to get it right? I have tried these two viewers: Volo View Express 2.01. On this page http://www.autodesk.no/adsk/index/0,,837403-123112,00.html it says: DWF-Only Version of Volo View Express. Views only DWF files and does not view DWG or DXF files. When I fill out the form and continue, the next page says: This version should understand DWG, DXFT, and DWF (ePlot and eView) files. What is right? WHIP! 4.0 On this page http://www.autodesk.com/cgi-bin/whipreg.pl it says: you can view and print AutoCAD drawings without using AutoCAD. On the next page a plug-in is downloaded and I can see a .dwf file in my browser. Still, when I try to execute my code to view drawing1.dwg it won't work. Is my code right, and if it is where can I download a .dwg-plug-in. regards, Lars Dave Merritt [EMAIL PROTECTED] skrev i melding news:109DB0BF6260D211A1B30008C7A4AA1B129674A4;postoffice.arvinmeritor.com... Lars, You say you downloaded the Volo View Express, but your plug-in page points to the WHIP viewer plug. Regardless, there are two versions of Volo View. Did you download the DWF version only of Volo View? Dave -Original Message- From: Lars Espelid [mailto:lars_espelid;hotmail.com] Sent: Friday, November 15, 2002 4:09 PM To: [EMAIL PROTECTED] Subject: [PHP] How to show autocad (dwf, dwg) files in browser? HI, I have downloaded Volo View Express from www.autodesk.com (plugin to web-browser) and can see the sample autocad-file that is presented on their site. Code ment to display the file drawing1.dwg on my page (none working): 1) img src=drawing1.dwg Displays no image. 2) OBJECT data=drawing1.dwg type=image/vnd.dwg A nice drawing. /OBJECT The only thing I can see is a blue rectangle whith the following error message inside it: Drawing File Format Unrecogr 3) object param name=Filename value=drawing1.dwg embed name=thanks src=drawing1.dwg pluginspage=http://www.autodesk.com/whip; /object The only thing I can see is a blue rectangle whith the following error message inside it: Drawing File Format Unrecogr I'm running Apache 1.3.26. This is what I have written into httpd.conf: AddType model/vnd.dwf .dwf #drawing/x-dwf .dwf AddType image/vnd.dxf .dxf AddType image/vnd.dwg .dwg This is what I have written into mime.types and mime.types.default: image/vnd.dwg image/vnd.dxf Appreciate any help. Thanks Lars -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Getting users IP address into a variable.
Try: $ip=$_SERVER['REMOTE_ADDR'] -Original Message- From: Webmaster MBTRADINGCO [mailto:[EMAIL PROTECTED]] Sent: Wednesday, October 02, 2002 1:08 PM To: [EMAIL PROTECTED] Subject: [PHP] Getting users IP address into a variable. I'm sure there has to be a way to verify which IP address is accessing from. I need to establish a page where when I enter it records the IP address I'm logging in from, to a database. Problem is I can't seem a command in php that can assign that to a variable, as in: $ip=HTTP_GET_ ANY IDEAS Thanks Elliot J. Balanza -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php ** Any views, opinions or authorizations contained in this email are solely those of the author and do not necessarily represent those of ArvinMeritor, Inc. If you are not familiar with the corporate authority of the author, please obtain confirmation in writing of the content of this email prior to taking any action on the basis of the information. If you are not the intended recipient, you are hereby notified that any disclosure, copying or distribution of the information enclosed is strictly prohibited. ** -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Creating/displaying page content and downloading a file at the same time
All, I'm trying to write a script that will allow the user to download/save a file but I also want to be able to display a web page in the browser window as well. If a user runs the following code from an empty browser window, then the contents of the browser window will remain empty. How do I get the script to also output page content to the browser window as well as download the file? I've played around with trying to add an additional text/html content type and echoing out text in different places in the script, but I cannot get any output to appear in the browser window. I've tried setting the location so that the page will open another page but then I run into the headers already sent problem. from the browser: http://myhost/download.php/?Filename=/tmp/20020923172330.xls download.php: ? $Filename = $_GET['Filename']; header(Content-Type: application/force-download); header(Content-Type: application/octet-stream); header(Content-Type: application/download); header(Content-Disposition: inline; filename= . basename($Filename)); header(Content-Transfer-Encoding: binary); header(Content-Length: . filesize($Filename)); header(Pragma: public); header(Expires: 0); header(Cache-Control: must-revalidate, post-check=0, pre-check=0); readfile($Filename); ? Thanks Dave Merritt [EMAIL PROTECTED] ** Any views, opinions or authorizations contained in this email are solely those of the author and do not necessarily represent those of ArvinMeritor, Inc. If you are not familiar with the corporate authority of the author, please obtain confirmation in writing of the content of this email prior to taking any action on the basis of the information. If you are not the intended recipient, you are hereby notified that any disclosure, copying or distribution of the information enclosed is strictly prohibited. ** -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Closing a browser's child window after user downloads a file
All,
I have a page that displays results of a database search. The page contains
a select dropdown box from which the user can pick different options
including the ability to download an Excel or CSV file of the displayed
results. Once the user selects an option from the dropdown, a Javascript
onchange method will kick in, submit the form and process the selected
option. If the option is one of the file download choices, then using
Javascript a new, child window will be opened and the location of the child
will be the PHP download script with the file to download. The script in
the child window will force the browser's download dialog box to open up at
which point the user can save, open, or cancel the file download. The issue
that I have is that some browsers (IE6) will automatically close the child
window when the user has made a selection from the download dialog box while
other browsers (Mozilla) will leave the child window remaining open. I need
to either come up with a way to close the child window once the user has
finsished with the download dialog or force the download script to not run
in a child window. I've thought about applying a timer somewhere but this
isn't reliable because of the unknowns of file download time and user
response time to the download dialog. I don't have a preference as to
whether the solution is all PHP based, all Javascript, or a mix of both -- I
just need a solution to close this child window when complete.
Code is below. As usual, thanks in advance.
Dave Merritt
[EMAIL PROTECTED]
Javascript code from the parent search results window opening the child
window calling the PHP download script:
...
SCRIPT type= text/javascript language=JavaScript
!--
DownloadWindow = window.open(../functions/download.php/?Filename=?echo
$Filename?, Download, toolbar=0, location=0, directories=0, status=0,
menubar=0, scrollbars=0, resizable=0, copyhistory=0, width=200,
height=200);
//--
/SCRIPT
...
download.php:
if (isset($_GET['Filename']))
{
if ( $_GET['Filename'] != '' )
{
$Filename = $_GET['Filename'];
header(Content-Disposition: inline; filename= .
basename($Filename));
header(Content-Type: application/octet-stream);
header(Content-Type: application/force-download);
header(Content-Type: application/download);
header(Content-Transfer-Encoding: binary);
header(Content-Length: . filesize($Filename));
header(Pragma: public);
header(Expires: 0);
header(Cache-Control: must-revalidate, post-check=0, pre-check=0);
readfile($Filename);
}
}
**
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author and do not necessarily represent those of ArvinMeritor, Inc. If you are not
familiar with the corporate authority of the author, please obtain confirmation in
writing
of the content of this email prior to taking any action on the basis of the
information. If
you are not the intended recipient, you are hereby notified that any disclosure,
copying
or distribution of the information enclosed is strictly prohibited.
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[PHP] OT - SQL string to get value by latest date only in a join
All, I apologize up front for being off topic, but I don't want to have to subscribe to other lists unless necessary. I know that someone on this list should be able to help me out. I have the SQL string below that I am running. The problem I'm having is that the history table being called in the join has multiple entries referencing back to the single entry in the customer table. What I want to be able to do is return only the latest/newest entry by date in the history table and not older entries. I assume that I need to add some conditional to the outer join. The problem I have is that I don't know what the conditional would be to allow returning the latest row by date. Thanks in advance, and again I apologize for the OT question, Dave Merritt [EMAIL PROTECTED] currently running this string: SELECT customer.customer_id, customer.customer_name, business_unit.business_unit, customer.created_date, CONCAT_WS(, , username.last_name, username.first_name), history.modify_date, CONCAT_WS(, , modified.last_name, modified.first_name), history.description FROM `customer` LEFT OUTER JOIN `business_unit` ON business_unit.bu_id = customer.bu_id LEFT OUTER JOIN `username` ON username.user_id = customer.creator_id LEFT OUTER JOIN `history` ON ( history.item_id = customer.customer_id AND history.module_id = 1003 ) LEFT OUTER JOIN `username` AS modified ON modified.user_id = history.modifier_id WHERE customer.customer_id LIKE %man% OR customer.customer_name LIKE %man% OR customer.created_date LIKE %man% OR business_unit.business_unit LIKE %man% OR username.first_name LIKE %man% OR username.last_name LIKE %man% OR history.modify_date LIKE %man% OR history.description LIKE %man% OR modified.first_name LIKE %man% OR modified.last_name LIKE %man% ORDER BY customer.customer_name which returns the following results: '1040','Allmand Bros. Inc.','Gas Springs, Industrial','2002-03-01 00:00:00','Admin, PDMWeb','2002-11-12 10:53:23','Burnstingle, Robert','Additional changes made','1003', '1040','Allmand Bros. Inc.','Gas Springs, Industrial','2002-03-01 00:00:00','Admin, PDMWeb','2002-11-09 13:15:34','Merritt, Dave','A test of history','1003', '1050','Ameritool Manufacturing','Gas Springs, Industrial','2002-03-01 00:00:00','Admin, PDMWeb','NULL','','NULL','NULL', '1053','Ammann-Yanmar','Gas Springs, Industrial','2002-03-01 00:00:00','Admin, PDMWeb','NULL','','NULL','NULL', '1065','Art's Way Manufacturing','Gas Springs, Industrial','2002-03-01 00:00:00','Admin, PDMWeb','NULL','','NULL','NULL', . What I'm trying to achieve is the above results both with only one row returned for the 1040 item like so, and the row returned should be the row with the latest/newest modified date: 1040','Allmand Bros. Inc.','Gas Springs, Industrial','2002-03-01 00:00:00','Admin, PDMWeb','2002-11-12 10:53:23','Burnstingle, Robert','Additional changes made','1003', '1050','Ameritool Manufacturing','Gas Springs, Industrial','2002-03-01 00:00:00','Admin, PDMWeb','NULL','','NULL','NULL', '1053','Ammann-Yanmar','Gas Springs, Industrial','2002-03-01 00:00:00','Admin, PDMWeb','NULL','','NULL','NULL', '1065','Art's Way Manufacturing','Gas Springs, Industrial','2002-03-01 00:00:00','Admin, PDMWeb','NULL','','NULL','NULL', . ** Any views, opinions or authorizations contained in this email are solely those of the author and do not necessarily represent those of ArvinMeritor, Inc. If you are not familiar with the corporate authority of the author, please obtain confirmation in writing of the content of this email prior to taking any action on the basis of the information. If you are not the intended recipient, you are hereby notified that any disclosure, copying or distribution of the information enclosed is strictly prohibited. ** -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Trying to locate an image file name from another site
All,
I have a page on our intranet site that is pulling an image from our
corporate web server. The corporate server the image I am accessing is
beyond my control. The image is generated daily and appears to be named
with a timestamp in the file name so therefore the image name changes daily.
On my site I have the page to check if a cookie exists with the image's file
name. If the cookie is set then the page will create the IMAGE tag using
the file name specified in the cookie. If the cookie value is not set then
the code will go open the web page on the corporate server, dump the page's
content into an array, and then the array is parsed until a known string is
found and then the image file name can be pulled out and displayed.
Currently the code I have is working. However, I know that it can/needs to
be improved but I'm not sure what/how to go about implementing the following
improvements:
1) If the image file name doesn't exist go to the corporate web server,
access the page, and parse the page contents until we find the image. The
problem I currently have is that the page on the corporate site is huge with
several hundred lines of html in the page and the image I need to access is
closer to the bottom of the page. How do I walk through the contents of the
page searching for the image starting further down the page than from the
beginning if the page?
2) Is there a better way of trying to find/strip the image name in the page
contents than by using the strpos substr functions? If so how?
3) Once I've found the image file name, is there then some way that I can
download the image from the corporate site to our local site? If so, how do
I do this? If I can download the image locally, then I can rewrite the code
to check for the existence of the file locally before going out and
accessing the corporate site every time.
Code is below. Thanks in advance
Dave Merritt
[EMAIL PROTECTED]
%
$Image = $_COOKIE['ImageToday'];
$WebLocation = 'http://www.somesite.com';
if (empty($Image))
{
$Find = 'IMG SRC=/archive/images';
$PageContents = @file($WebLocation);
if (! empty($PageContents))
{
while (list ($ArrayNo, $Line) = each ($PageContents))
{
$Image = stristr($Line, $Find);
if ($Image != )
{
$Image = substr($Image, 0, strpos($Image, ' BORDER=0'));
$Image = substr($Image, strpos($Image, '') + 1);
$Image = $WebLocation . $Image;
%
SCRIPT type= text/javascript language=JavaScript
!--
document.cookie = ImageToday=%echo $Image%; path=/
//--
/SCRIPT
A href=%echo $WebLocation%
IMG src=%echo $Image% border=0 alt=Some label
/A
%
}
}
}
else
{
%
P class=textUnable to display
A href=%echo $WebLocation% class=textimage from
site/A!
/P
%
}
}
else
{
%
A href=%echo $WebLocation%IMG src=%echo $Image% border=0
alt=Some label/A
%
}
%
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RE: [PHP] Trying to locate an image file name from another site
No. Having any changes made to the corporate server for my ease of use is not an option -- too much politics involved (my use of open source solutions in a Microsoft environment!!!). Dave -Original Message- From: Miguel Cruz [mailto:[EMAIL PROTECTED]] Sent: Tuesday, July 09, 2002 3:55 PM To: Merritt, Dave Cc: PHP General (E-mail) Subject: Re: [PHP] Trying to locate an image file name from another site On Tue, 9 Jul 2002, Merritt, Dave wrote: I have a page on our intranet site that is pulling an image from our corporate web server. The corporate server the image I am accessing is beyond my control. The image is generated daily and appears to be named with a timestamp in the file name so therefore the image name changes daily. Can you ask the corporate web site people to insert a distinctive HTML comment just before the image? It would just take them a second, have no impact on their users, and make your job much easier. miguel -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
